# LCM & HCF

## LCM & HCF

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Question 1 |

The LCM of two numbers is 936. If their HCF is 4 and one of the numbers is 72, the other is ?

A | 42 |

B | 52 |

C | 62 |

D | None of these |

Question 1 Explanation:

Use HCF × LCM = product of numbers.

Question 2 |

Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for the first time?

A | 12:10 P.M. |

B | 12:12 P.M. |

C | 12:11 P.M. |

D | None of these |

Question 2 Explanation:

The first bell toll will be after the time elapsed is the LCM of 50 and 48. The LCM of 50 and 48 is 50 × 24 = 1200. Hence, the first time they would toll together after 12 noon would be exactly 1200 seconds or 20 minutes later.

Question 3 |

4 Bells toll together at 9:00 A.M. They toll after 7, 8, 11 and 12 seconds respectively. How many times will they toll together again in the next 3 hours?

A | 3 |

B | 4 |

C | 5 |

D | 6 |

Question 3 Explanation:

The LCM of 7, 8, 11 and 12 is given by 12 × 11 × 2 × 7 = 264 × 7 = 1848. 1848 seconds is 30 minutes 48 seconds. Hence, the 4 bells would toll together every 30 minutes 48 seconds.
The number of times they would toll together in the next 3 hours would be given by the quotient of the division: 3 × 60 × 60 /1848 =5 times

Question 4 |

On Ashok Marg three consecutive traffic lights change after 36, 42 and 72 seconds respectively. If the lights are first switched on at 9:00 A.M. sharp, at what time will they change simultaneously?

A | 9 : 08 : 04 |

B | 9 : 08 : 24 |

C | 9 : 08 : 44 |

D | None of these |

Question 4 Explanation:

The LCM of 36,42 and 72 is 504. Hence, the lights will change simultaneously after 8 minutes and 24 seconds.

Question 5 |

The HCF of 2472, 1284 and a third number ‘N’ is 12. If their LCM is 2^3 × 3^3 × 5^1 × 103 × 107, then the number ‘N’ is: ?

A | 2^2 × 3^2 × 7^1 |

B | 2^2 × 3^3 × 103 |

C | 2^2 × 3^2 × 5^1 |

D | None of these |

Question 5 Explanation:

2472 = 2^3 × 103 × 3; 1284 = 2^2 × 107 × 3. Since the HCF is 12, the number must have a component of 2^2 × 3^1 at the very least in it. Also, since the LCM is 2^3 × 3^2 × 5^1 × 103 × 107
we can see that the minimum requirement in the required number has to be 3^2 × 5^1. Combining these two requirements we get that the number should have 2^2 × 3^2 × 5^1 at the minimum and the power of 2 could also be 2^3 while we could also have either one of 103^1 and/or 107^1 as a part of the required number.

Question 6 |

Two equilateral triangles have the sides of lengths 34 and 85 respectively. The greatest length of tape that can measure both of them exactly is: ?

A | 34 |

B | 85 |

C | 15 |

D | 17 |

Question 6 Explanation:

HCF of 34 and 85 is 17.

Question 7 |

Two numbers are in the ratio 17:13. If their HCF is 15, what are the numbers?

A | 255 and 190 |

B | 200 and 100 |

C | 255 and 195 |

D | 200 and 195 |

Question 7 Explanation:

17 × 15 and 13 × 15 i.e. 255 and 195 respectively.) [Note : This can be done when the numbers are co-prime.]

Question 8 |

A forester wants to plant 44 apple trees, 66 banana trees and 110 mango trees in equal rows (in terms of number of trees). Also, he wants to make distinct rows of trees (i.e. only one type of tree in one row). The number of rows (minimum) that are required are: ?

A | 2 |

B | 3 |

C | 10 |

D | 11 |

Question 8 Explanation:

44/22 + 66/22 + 110/22 (Since 22 is the HCF)

Question 9 |

Three runners running around a circular track can complete one revolution in 2, 4 and 5.5 hours respectively. When will they meet at the starting point ?

A | 22 |

B | 23 |

C | 11 |

D | 44 |

Question 9 Explanation:

The answer will be the LCM of 2, 4 and 11/2. This will give you 44 as the answer

Question 10 |

The HCF and LCM of two numbers are 33 and 264 respectively. When the first number is divided by 2, the quotient is 33. The other number is?

A | 66 |

B | 132 |

C | 198 |

D | 99 |

Question 10 Explanation:

33 × 264 = 66 × n. Hence, n = 132

Question 11 |

The greatest number which will divide: 4003, 4126 and 4249 ?

A | 43 |

B | 41 |

C | 45 |

D | None of these |

Question 11 Explanation:

Greatest number with which if we divide P,Q,R and it leaves same remainder in each case. Number is of form = HCF of (P-Q),(P-R).
Therefore, HCF of (4126 - 4003), (4249 - 4003) = HCF of 123, 246 = 41.[Taken for Positive result].
Detailed Explanation:
The numbers can be written as,
4003 = AX +P where P = Remainder
4126 = BX +P
4249 = CX +P
(B - A)*X = 123
(C - B)*X = 246
Thus the X is factor of 123 and 246.

Question 12 |

Which of the following represents the largest 4 digit number which can be added to 7249 in order to make the derived number divisible by each of 12, 14, 21, 33, and 54. ?

A | 9123 |

B | 9383 |

C | 8727 |

D | None of these |

Question 12 Explanation:

The LCM of the numbers 12, 14, 21, 33 and 54 is 8316. Hence, in order for the
condition to be satisfied we need to get the number as: 7249 + n = 8316 × 2
Hence, n = 9383.

Question 13 |

Find the greatest number of 5 digits, that will give us a remainder of 5, when divided by 8 and 9 respectively. ?

A | 99931 |

B | 99941 |

C | 99725 |

D | None of these |

Question 13 Explanation:

The LCM of 8 and 9 is 72. The largest 5 digit multiple of 72 is 99936. Hence, the
required answer is 99941.

Question 14 |

The least perfect square number which is divisible by 3, 4, 6, 8, 10 and 11 is: ?

A | 3 × 3 × 3 × 3 × 2 × 2 × 5 × 5 × 11 × 11 |

B | 3 × 3 × 2 × 2 × 2 × 2 × 1× 1 × 11 × 11 |

C | 7 × 7 × 2 × 2 × 2 × 2 × 5 × 5 × 11 × 11 |

D | 3 × 3 × 2 × 2 × 2 × 2 × 5 × 5 × 11 × 11 |

Question 14 Explanation:

The number should have at least one 3, three 2’s, one 5 and one 11 for it to be divisible by 3, 4, 6, 8, 10 and 11.
Further, each of the prime factors should be having an even power in order to be a perfect square.
Thus, the correct answer will be: 3 × 3 × 2 × 2 × 2 × 2 × 5 × 5 × 11 × 11

Question 15 |

Find the greatest number of four digits which when divided by 10, 11, 15 and 22 leaves 3, 4, 8 and 15 as remainders respectively. ?

A | 9907 |

B | 9903 |

C | 9893 |

D | None of these |

Question 15 Explanation:

First find the greatest 4 digit multiple of the LCM of 10, 11, 15 and 22. (In this case it is
9900). Then, subtract 7 from it to give the answer.

Question 16 |

Find the HCF of (3^125 – 1) and (3^35 – 1) ?

A | 10 |

B | 4 |

C | 7 |

D | 5 |

Question 16 Explanation:

The solution of this question is based on the rule that: The HCF of (am – 1) and (a^n – 1) is given by (a^(HCF of m, n) – 1)
Thus, in this question the answer is: (3^5 – 1). Since 5 is the HCF of 35 and 125.

Question 17 |

What will be the least possible number of the planks, if three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length?

A | 7 |

B | 8 |

C | 22 |

D | 11 |

Question 17 Explanation:

The least possible number of planks would occur when we divide each plank into a length equal to the HCF of 42, 49 and 63. The HCF of these numbers is clearly 7- and this should be the size of each plank. Number of planks in this case would be: 42/7 + 49/7 + 63/7 = 6 + 7 + 9 = 22 planks.

Question 18 |

Find the greatest number, which will divide 215, 167 and 135 so as to leave the same remainder in each case. ?

A | 64 |

B | 32 |

C | 24 |

D | 16 |

Question 18 Explanation:

Trial and error would give us that the number 16 would leave the same remainder of 7 in all the three cases.

Question 19 |

Find the L.C.M of 2.5, 0.5 and 0.175 ?

A | 2.5 |

B | 5 |

C | 7.5 |

D | 17.5 |

Question 19 Explanation:

The numbers are 5/2, 1⁄2 and 175/1000 = 7/40. The LCM of three fractions is given by the formula:LCM of numerators/HCF of denominators = (LCM of 5, 1 and 7)/(HCF of 2 and 40) = 35/2 =17.5

Question 20 |

The L.C.M of 4.5; 0.009; and 0.18 = ?
(a)(b)
(c) (d)

A | 4.5 |

B | 45 |

C | 0.225 |

D | 2.25 |

Question 20 Explanation:

Use the same process as for question 23 for the numbers: 9/2; 9/1000 and 9/50. (LCM of 9, 9, 9)/ (HCF of 2, 100, 50) = 9/2 = 4.5

Question 21 |

The L.C.M of two numbers is 1890 and their H.C.F is 30. If one of them is 270, the other will be ?

A | 210 |

B | 220 |

C | 310 |

D | 320 |

Question 21 Explanation:

1890 × 30 = 270 × N then N = 210.

Question 22 |

What is the smallest number which when increased by 6 is divisible by 36, 63 and 108 ?

A | 750 |

B | 752 |

C | 754 |

D | 756 |

Question 22 Explanation:

The LCM of 36, 63 and 108 is 756. Hence, the required number is 750.

Question 23 |

The smallest square number, which is exactly divisible by 2, 3, 4, – 9, 6, 18, 36 and 60, is ?

A | 900 |

B | 1600 |

C | 3600 |

D | None of these |

Question 23 Explanation:

The LCM of the given numbers is 180. Hence, all multiples of 180 would be divisible by all of these numbers. Checking the series 180, 360, 540, 720, 900 we can see that 900 is the first perfect square in the list.

Question 24 |

The H.C.F of two numbers is 11, and their L.C.M is 616. If one of the numbers is 88, find the other. ?

A | 77 |

B | 87 |

C | 97 |

D | None of these |

Question 24 Explanation:

Using the property HCF × LCM = product of the numbers, we get:
616 × 11 = 88 × N Then N= 77.

Question 25 |

What is the greatest possible rate at which a man can walk 51 km and 85 km in an exact number of minutes?
(a) (b)
(c) (d)

A | 11 km/min |

B | 13 km/min |

C | 17 km/min |

D | None of these |

Question 25 Explanation:

The answer would be given by the HCF of 51 and 85 – which is 17.

Question 26 |

The HCF and LCM of two numbers are 12 and 144 respectively. If one of the numbers is 36, the other number is ?

A | 4 |

B | 48 |

C | 72 |

D | 432 |

Question 26 Explanation:

Using the property HCF × LCM = product of the numbers, we get: 12 × 144 = 36 × N Then N = 48.

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